Ask question

Ask question

All categories

3 years ago

11

If the random variable x is normally distributed with a mean equal to .45 and a standard deviation equal to .40, then P(x ≥ .75)

is:

1 answer:

stellarik [79]3 years ago

4 0

P(x ≥ 0.75) = P(z ≥ (0.75 - 0.45)/0.40) = P(z ≥ 0.30/0.40) = P(z ≥ 0.75) = 1 - P(z < 0.75) = 1 - 0.077337 = 0.22663

You might be interested in

docker41 [41]

answer : option C

$f(x) = x^3 + x^2 - 8x - 8$

Lets find the number of zeros by factoring

$0 = x^3 + x^2 - 8x - 8$

Group first two terms and last two terms

$0 = (x^3 + x^2) (- 8x - 8)$

Factor out GCF from each group

$0 = x^2(x + 1)-8(x + 1)$

$0 = (x^2-8)(x + 1)$

Now we set each factor =0 and solve for x

0 = (x^2-8) and (x + 1)=0

$x^2 = 8$ and x = -1

$x= 2+-\sqrt{2}$ and x= -1

So we have three real zeros

The function has three real zeros. The graph of the function intersects the x-axis at exactly three locations.

8 0

4 years ago

Maggie is planning on going to Penn State University. She estimates that she will need to earn at least $2,000 to live off of ov

professor190 [17]

A. The amount of money she must earn in the summer must be
x ≥ 2450

B. The inequality that describes the situation is
30x ≤ 2000

C. Solving the inequality
x ≤ 66.67 ~ 66 weeks

D. The answer is Part C means that she can only make use of the $2000 money for food for 66 weeks before it's not enough.

8 0

3 years ago

A box of colored crayons contains 9 distinct colors. In how many ways can 2 colors be chosen, assuming that the order of the col

shepuryov [24]

Answer:

36

Step-by-step explanation:

We can use combinations

9 choose 2 =

9!/(7!*2!) = (9*8)/2 = 36

3 0

3 years ago

What is the greatest common factor of the terms in the expression 7/3ab minus 7/6b?

Cloud [144]

Answer:

7b, is the answer. they both have a numerator of 7, and the both have x

Step-by-step explanation:

7 0

4 years ago

Read 2 more answers

Please help me answer this question

stealth61 [152]

The value of dy/dx for the functions are

(i) $\frac{dy}{dx} = 4x^{2}sin2x.cos2x+ 2x. sin^{2}2x$

(ii) $\frac{dy}{dx} =\frac{- y(1+3x^{2})}{2x(1+x^{2}) }$

<h3>Differentiation</h3>

From the question, we are to determine dy/dx for the given functions

(i) $x^{2} sin^{2}2x$

Let $u = x^{2}$

and

$v = sin^{2} 2x$

Also,

Let $w= sin2x$

∴ $v = w^{2}$

First, we will determine $\frac{dv}{dx}$

Using the Chain rule
$\frac{dv}{dx} = \frac{dv}{dw}.\frac{dw}{dx}$

$v = w^{2}$

∴ $\frac{dv}{dw} =2w$

Also,

$w= sin2x$

∴ $\frac{dw}{dx} =2cos2x$

Thus,

$\frac{dv}{dx} = 2w \times 2cos2x$

$\frac{dv}{dx} = 2sin2x \times 2cos2x$

$\frac{dv}{dx} = 4sin2x . cos2x$

Now, using the product rule

$\frac{dy}{dx} = u\frac{dv}{dx} + v\frac{du}{dx}$

From above

$u = x^{2}$

∴ $\frac{du}{dx}=2x$

Now,

$\frac{dy}{dx} = x^{2} (4sin2x.cos2x)+ sin^{2}2x (2x)$

$\frac{dy}{dx} = 4x^{2}sin2x.cos2x+ 2x. sin^{2}2x$

(ii) $xy^{2}+y^{2}x^{3} +2=0$

Then,

$x.2y\frac{dy}{dx}+ y^{2}(1)+y^{2}.3x^{2} + x^{3}.2y\frac{dy}{dx} +0=0$

$2xy\frac{dy}{dx}+ y^{2}+3x^{2}y^{2} + 2x^{3}y\frac{dy}{dx} =0$

$2xy\frac{dy}{dx}+2x^{3}y\frac{dy}{dx} =- y^{2}-3x^{2}y^{2}$

$\frac{dy}{dx} (2xy+2x^{3}y)=- y^{2}(1+3x^{2})$

$\frac{dy}{dx} =\frac{- y^{2}(1+3x^{2})}{2xy+2x^{3}y}$

$\frac{dy}{dx} =\frac{- y^{2}(1+3x^{2})}{2xy(1+x^{2}) }$

$\frac{dy}{dx} =\frac{- y(1+3x^{2})}{2x(1+x^{2}) }$

Hence, the value of dy/dx for the functions are

(i) $\frac{dy}{dx} = 4x^{2}sin2x.cos2x+ 2x. sin^{2}2x$

(ii) $\frac{dy}{dx} =\frac{- y(1+3x^{2})}{2x(1+x^{2}) }$

Learn more on Differentiation here: brainly.com/question/24024883

#SPJ1

8 0

2 years ago